# -*- coding: utf-8 -*-#
#-------------------------------------------------------------------------------
# 建立者:        黄周云  
# Name:         test10
# Description:  例7.7
# Author:       ASUS
# Date:         2019/3/9
#-------------------------------------------------------------------------------

# 递归深度为1000
# 计算阶乘的递归函数
def jc(n):
    if n <= 0 :
        return 1
    else:
        return n * jc(n-1)
print(jc(10))

# 计算斐波那契数列的递归函数
def fibonaccci(n):
    if n == 1:
        return 0
    elif n == 2:
        return 1
    else:
        return fibonaccci(n-1)+fibonaccci(n-2)
# 斐波那契数列的第十个值
print(fibonaccci(10))


